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Suppose we have a function for the velocity given by $v(r,\theta)=r$, or in Cartesian form $v(x,y)=\sqrt{x^2+y^2}$. As we can see below, as we get closer to the origin $(0,0)$, the velocity decreases. I've found it easy to visualize the field as being some form of quicksand where it is harder to move through as you approach the origin.

This is demonstrated below by a plot I've made using Wolfram Mathematica: enter image description here

What I am trying to do: Find the two functions for $y(x)$ which would minimize the time taken to get from point $A(-1,0)$ and $B(1,0)$.

I deduced that the fastest path cannot be the straight line directly from $A$ to $B$, since it would require an infinite time to get through the origin. A guess for the two curves is shown by the $\color{#0050B0}{\text{dark blue}}$ and the $\color{#00AAAA}{\text{light blue}}$ curves I've made. I'm almost certain they would be symmetrical. I've first guessed that the optimized curve would be similar to an ellipse, however I hesitated after I've plotted this.

I've done some research on the problem and figured it may be similar to the derivation of the Brachistochrone curve, using the Euler-Lagrange equations.

I am new to the Calculus of Variations, so here is the working I've done so far.

We have:

$$dt=\frac{ds}{v} \Rightarrow dt=\frac{\sqrt{dx^2+dy^2}}{\sqrt{x^2+y^2}} \Rightarrow dt=\frac{\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}}{r}~d\theta$$

On the third step I converted it to polar coordinates. Adding integration signs:

$$\int_{0}^{T}~dt=\int_{\theta_1}^{\theta_2}\frac{\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}}{r}~d\theta$$

$$T=\int_{\theta_1}^{\theta_2} \sqrt{1+\frac{(r')^2}{r^2}}~d\theta$$

Where $T$ is the total time taken to get from $A$ to $B$. I thought of using the following Euler-Lagrange Equation:

$$\frac{d}{d\theta}\left(\frac{\partial L}{\partial r'}\right)=\frac{\partial L}{\partial r} \tag{1}$$ For the functional: $$L(\theta,r,r')=\sqrt{1+\frac{(r')^2}{r^2}}$$ Evaluating partial derivatives:

$$\frac{dL}{dr}=-\frac{(r')^2}{r^3\sqrt{\frac{(r')^2}{r^2}+1}}=-\frac{(r')^2}{r^2\sqrt{(r')^2+r^2}}$$ $$\frac{dL}{dr'}=\frac{r'}{r^2\sqrt{\frac{(r')^2}{r^2}+1}}=\frac{r'}{r\sqrt{(r')^2+r^2}}$$

Substituting into $(1)$:

$$\frac{d}{d\theta}\left(\frac{r'}{r\sqrt{(r')^2+r^2}}\right)=-\frac{(r')^2}{r^2\sqrt{(r')^2+r^2}}$$

I integrated both sides with respect to $\theta$ and obtained:

$$\frac{r'}{r\sqrt{(r')^2+r^2}}=-\frac{(r')^2\theta}{r^2\sqrt{(r')^2+r^2}}+C \tag{2}$$

Now, I realize I must solve this differential equation. I've tried simplifying it to obtain:

$$r\frac{dr}{d\theta}=-\left(\frac{dr}{d\theta}\right)^2\theta+Cr^2\sqrt{\left(\frac{dr}{d\theta}\right)^2+r^2} \tag{3}$$

However, I think I've hit a dead end. I'm not certain that it is solvable in terms of elementary functions. Both Mathematica and Wolfram|Alpha have not given me a solution to this differential equation.

To conclude, I would like some guidance on how to continue solving the differential equation, assuming I have done the calculation and methodology correctly so far. If I have not done the correct methodology, I would appreciate some guidance on how to proceed with the problem.

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    Well one obvious solution to your differential equation is when $r$ is constant. Have you looked at that? – 2017-02-04
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    I did a simulation where movement occurs along a straight line from point A to a point above the center $(0,y_1)$ and then from there along a straight line to point B. The fastest path occurs when $y_1=1$, where the time is $t=3.54$. This is still worse than the time it takes to simply move along a circle from A to B, which takes $t=\pi$. I very much suspect that the optimal path is the cirular path. – 2017-02-04
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    Note that your step where you "integrated both sides with respect to $\theta$" is invalid: $r$ and $r'$ can depend on $\theta$, and so $\int f(r(\theta), r'(\theta)) d\theta \neq f(r(\theta), r'(\theta))\theta$. – 2017-04-08

2 Answers 2

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I think we can do without variational calculus.

Write the curve going from $B$ to $A$ in the form $$\gamma:\quad\theta\mapsto{\bf z}(\theta):=e^{u(\theta)}(\cos\theta,\sin\theta)\qquad(0\leq\theta\leq\pi)\ .$$Then $$ds=\sqrt{r^2+r'^2}\>d\theta=\sqrt{1+u'^2(\theta)}\>e^{u(\theta)}\>d\theta\ ,$$ hence $$dt={ds\over v\bigl({\bf z}(\theta)\bigr)}=\sqrt{1+u'^2(\theta)}\>d\theta\ .$$ It follows that we should minimize $$T=\int_0^\pi\sqrt{1+u'^2(\theta)}\>d\theta$$ under the condition $u(0)=u(\pi)=0$. It is obvious that $u(\theta)\equiv0$ does the job.

For general boundary data $A$, $B$ the pullback $\hat \gamma$ in the $(\theta,u)$-plane has to be a straight line, hence $\gamma$ will be an arc of a logarithmic spiral.

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    Thanks for answering! I've verified this by solving the Euler-Lagrange equations myself from the functional you've obtained: $$T=\int_0^{\pi} \sqrt{1+{u'}^2(\theta)}~d\theta$$ And got a parametric curve in the form: $$x(\theta)=ae^{b\theta}\cos{\theta}$$ $$y(\theta)=ae^{b\theta}\sin{\theta}$$ in the general case, which is indeed a logarithmic spiral. When I substituted the initial conditions I obtained a unit circle as you've implicitly mentioned. – 2017-02-04
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Just to show that this can be done using the calculus of variations: start with your functional $$ L(\theta,r,r')=\sqrt{1+\frac{(r')^2}{r^2}} $$ Now, we have that $\partial L/\partial \theta = 0$, which implies (via the Beltrami identity) that the quantity $$ L - r' \frac{\partial L}{\partial r'} = C $$ where $C$ is a constant with respect to $\theta$. In your case, this implies that $$ C = \sqrt{1+\frac{(r')^2}{r^2}} - r' \frac{r'/r^2}{\sqrt{1+\frac{(r')^2}{r^2}}} = \frac{1}{\sqrt{1+\frac{(r')^2}{r^2}}} $$ Re-arranging, we find that $$ \frac{r'}{r} = \sqrt{\frac{1}{C^2} - 1} $$ which itself is another constant $D$; thus, we have $r' = D r$, or $r = e^{D\theta}$ for some constant $D$.